I'm having problems dividing the polynomial 2x^5 + 2x^3 + x^2 + 1 by x^2 + 1

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- Mar 22nd 2011, 10:04 AMDevi09Dividing polynomial
I'm having problems dividing the polynomial 2x^5 + 2x^3 + x^2 + 1 by x^2 + 1

- Mar 22nd 2011, 10:19 AMHallsofIvy
Looks straight forward to me. But I don't know what to tell you without seeing what you did. Did you remember to include "0x" in the divisor and "$\displaystyle 0x^4$" and "$\displaystyle 0x$" in the dividend? How many times does $\displaystyle x^2$ divide into $\displaystyle 2x^5$?

- Mar 22nd 2011, 10:39 AMDevi09
It goes in 2x^3 times but I get x^2 as the next quotient part but the correct quotient should be 2x^3 - 1 according to the answer the book gives

- Mar 22nd 2011, 03:03 PMDevi09
$\displaystyle x^2+0x+1sqrt(2x^5+0x^4+2x^3+x^2+0x+1)$

This is the subtracting parts: The quotient i got is $\displaystyle 2x^3+x^2+x+1$

Part 1 $\displaystyle 2x^5+0x^4+2x^3$

Part 2 $\displaystyle x^4+x^2$

$\displaystyle x^4+x^3+x^2$

Part 3 $\displaystyle x^3+0x$

$\displaystyle x^3+x^2+x$

Part 4 $\displaystyle x^2+1$

$\displaystyle x^2+x+1$

Final $\displaystyle x$ - Mar 23rd 2011, 04:54 AMHallsofIvy
Yes, $\displaystyle x^2$ divides into $\displaystyle 2x^5$ $\displaystyle 2x^3$ times. And multiplying the entire divisor by $\displaystyle 2x^3$ gives $\displaystyle 2x^5+ 2x^3$

Now subtract that from the dividend:

$\displaystyle 2x^5+ 2x^3+ x^2+ 1- (2x^5+ 2x^3)= x^2+ 1$.

And now, of course, $\displaystyle x^2+ 1$ divides into that exactly once: the quotient is $\displaystyle 2x^3+ 1$.